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Verify the identity algebraically:
Csc(-x)tanx =-secx


Sagot :

Step-by-step explanation:

Recall that

[tex]\sin(-x) = -\sin x[/tex]

Therefore,

[tex]\csc(-x) = \dfrac{1}{\sin(-x)} = -\dfrac{1}{\sin x}[/tex]

so

[tex]\csc(-x)\tan x = \left(-\dfrac{1}{\sin x}\right)\left(\dfrac{\sin x}{\cos x}\right)[/tex]

[tex]\:\:\:\:\:\:\:\:\:= -\dfrac{1}{\cos x} = -\sec x[/tex]